An Introduction to Thermodynamics phần 3 docx

without changing their effect, and therefore decreases the pressure on the walls, requiring the “supplement” to P to fit the ideal gas equation form; see Phys. Teach. 34, 248-249 (April, 1996). The reader with some knowledge of special relativity may recognize that the total 16 energy of any system is measured by its mass, multiplied by the square of the speed of light. However, it would be necessary to measure masses about a million times more accurately than is now possible to be able to determine energies to the accuracy required in thermochemistry. 7/10/07 1- 27 It should also be possible to relate the volume of any liquid or solid to its temperature and pressure, or to express such other properties as refractive index, heat capacity at constant volume or pressure, thermal conductivity, heats of vaporization or fusion, or vapor pressures of solids or liquids, in terms of the temperature and pressure. Some of these equations will be encountered in later chapters. Thermochemistry The application to chemical reactions of the principles developed thus far is called thermochemistry. In particular, the heats of reaction are measured and tabulated and from these and from measured heat capacities the enthalpy changes are calculated for other reactions or for other experimental conditions. HESS’S LAW. The enthalpy change for a chemical reaction, such as the oxidation of sulfur dioxide to sulfur trioxide — 2 SO (g) + O &6 2 SO (liq) 2 2 3 — can be expressed as the difference between the enthalpies of the initial and final states. ∆H = H - H reaction final initial = H(2 SO ) - H(2 SO ) - H(O ) 3 2 2 There is no way within thermodynamics of measuring an absolute energy, or an absolute 16 enthalpy. Only energy, and enthalpy, changes can be determined. However, knowing that these energy and enthalpy changes depend only on the initial and final states, it is possible to add and subtract chemical reactions and add and subtract the corresponding enthalpy changes. That is, we may quite arbitrarily select a reference energy and/or enthalpy level and measure all values from that arbitrary level. In particular, it is possible to tabulate “heats of formation”, the enthalpy changes in the reaction of the elements to form each compound, and from these to calculate enthalpies of other reactions. This principle is known as Hess’s law. The reactions for the formation of the gases SO and SO from the elements are 2 3 S + O SO 2 2 S + 3/2 O SO (liq) 2 3 7/10/07 1- 28 The measured enthalpy changes for these reactions at 25 C and 1 atm pressure are -296.90 kJ/mol o and -437.94 kJ/mol. Subtraction of the first reaction from the second gives SO + ½ O &6 SO (liq) 2 2 3 and subtraction of the enthalpy changes gives -141.04 kJ/mol, which is the heat of reaction for the oxidation of SO to SO (liq). 2 3 Exactly the same elements, in the same quantities, always appear on both sides of a chemical equation (which is why reactions as written are called “equations”). Subtraction of the elements from both sides of an equation will yield, on each side, product minus reactants for the reactions of formation of each of the substances appearing in the original equation. In the example above, the original equation was SO + ½ O &6 SO . Subtract 1 mol of S and 3/2 mol of O from each 2 2 3 2 side. The equation can then be written as the formation of each compound (i.e., of SO , O , and 2 2 SO ) from the elements. 3 (SO - S - O ) + (½ O - ½ O ) &6 (SO - S - 3/2 O ) 2 2 2 2 3 2 and therefore ∆H = ∆H (SO ) - ∆H (SO ) - ∆H (½ O ) reaction form form form 3 2 2 = -437.94 kJ/mol -296.90 kJ/mol - 0 = -141.04 kJ/mol(SO liq) 3 (Notice that the heat of formation of any element, in its standard state, is necessarily zero.) An entirely equivalent way of obtaining the same numbers is to consider the enthalpy of each compound on a scale taken with reference to the elements. Such enthalpy values are called standard enthalpies of the compounds; they are identical with the standard enthalpies of formation. Hess’s law can often be applied to find heats of reaction that could not be directly measured experimentally. For example, the reaction of two molecules of ethylene, C H , to form 2 4 cyclobutane, C H , would not readily occur quantitatively under conditions conducive to 4 8 measurement of the heat of reaction. But both ethylene and cyclobutane can be burned in oxygen, and subtraction of these reactions gives the reaction equation desired. 2 C H + 8 O &6 4 CO + 4 H O 2 4 2 2 2 C H + 8 O &6 4 CO + 4 H O 4 8 2 2 2 Subtraction of the second from the first gives 2 C H &6 C H 2 4 4 8 and, therefore, subtraction of the ∆H for the second combustion from the ∆H for the first combustion gives ∆H for the condensation reaction. Heats of combustion (equal to - ∆H ) reaction are comparatively easy to measure and are often tabulated. KIRCHHOFF’S LAW. The heat of reaction at a temperature other than that given in a table can be ( ) ( ) ∫ ∫ +∆+=∆ 1 2 2 1 products1reactants2 (17) T T T T PP dTCHdTCH ( ) [ ] ∫ −+∆=∆ 2 1 ) reactants(products12 T T PP dTCCHH 7/10/07 1- 29 found by calculating enthalpy changes along an arbitrary path. The total enthalpy change is independent of this choice of path. The method is known as Kirchhoff’s law. Assume that ∆H is known for a reaction at a temperature T and the ∆H at another temperature, 1 T , is to be found. Starting with the hot reactants at T (Figure 4), the reaction could be carried 2 2 out isothermally to obtain products at the same temperature. An alternative path would be to cool the reactants to the temperature T , carry out the reaction isothermally at T , and warm the 1 1 products to T . The heat of reaction at T is already known and if the heat capacities at constant 2 1 pressure are known, the enthalpy changes can be calculated for the processes of cooling reactants and warming products. This path must give the same ∆H as the isothermal reaction at T . 2 or, because interchanging limits of an integral will change the sign, If the difference in heat capacities is independent of temperature, this may be rewritten in the form ∆H = ∆H + [C - C ](T - T ) (18) 2 1 P 2 1 p (products) (reactants) For example, given that the heat of reaction for rhombic sulfur burning in oxygen to yield sulfur dioxide gas is - 296.9 kJ/mol at 25 C (298 K), find ∆H at 95 C (368 K). The heat o o capacities are given in Table 2. Insertion of the numerical values into equation 18 gives Table 2 HEAT CAPACITIES ' Average values (in J/mol-K) for temperature ranges indicated Compound C Temperature, C P o He 20.8 -200 up H 28.8 25 to 200 2 O 29.4 25 to 200 2 H O(g)* 36.4 25 to 200 2 SO (g) 41.9 25 to 200 2 S (r) 23.7 25 to 200 S (m) 25.9 95 to 120 *For rough calculations it is sufficient to set C (steam) = C (ice) = ½ C (liq H O). P P P 2 This property associated with state functions has sometimes been confused with a 17 conservation principle. Enthalpy is not conserved. 7/10/07 1- 30 ∆H = - 296,900 J/mol + (41.9- 29.4 - 23.7) x 70 J/mol 368 = - 296.1 J/mol Sometimes there will be a phase transition during the warming or cooling process. Sulfur has a phase change at 95 C, at which point rhombic sulfur goes to monoclinic sulfur; the monoclinic o sulfur melts at 119 C. The enthalpy changes are 11.78 and 39.24 kJ/mol. The heat of reaction o for liquid sulfur burning in oxygen to form SO at 119 C (392 K) can be calculated as follows (see 2 o Figure 5). ∆H = - ∆H - C (m) (119 - 95) - ∆H - C (r) (95 - 25) - C (O ) (119 - 25) 392 fusion P tr P P 2 + ∆H + C (SO ) (119 - 25) 298 P 2 ∆H = - 39,240 - 25.9 x 24 - 11,780 - 23.7 x 70 - 29.4 x 94 - 296,900 + 41.9 x 94 J/mol 392 = - 349.0 kJ/mol Note that temperature differences can be found without conversion to the Kelvin scale. Both Hess’s law and Kirchhoff’s law are simply applications of the principle that changes in a state function, such as the enthalpy, are completely determined by the initial and final states. 17 This principle is combined with the equation arising from the first law that shows that i f the pressure is constant, the enthalpy change will be equal to the heat absorbed by the system. Thus the “heat of reaction,” by which we mean ∆H (at a particular temperature, pressure, and reaction concentrations of reactants and products), is only equal to the heat absorbed i f the reaction proceeds at constant pressure (and at the specified temperature and concentrations). It is sometimes more convenient to carry out a reaction at constant volume. Then the heat absorbed is not equal to the “heat of reaction” (that is, to ∆H ), but it is still determinate because heat reaction absorbed equals the change in energy when the system follows a constant-volume path and because ∆E is fixed by the initial and final states. reaction 7/10/07 1- 31 The experimental determination of a heat of reaction is called calorimetry. A typical calorimeter (Figure 6) consists of a reaction chamber, surrounded by a layer of water, enclosed by sufficient insulation to prevent heat loss to the surrounding laboratory. The reactants, at room temperature, are placed in the reaction chamber, the calorimeter is closed, and the reaction is initiated by an electrically heated wire or other controlled energy source. The reaction will normally be exothermic and the reaction chamber will therefore become quite hot, but the heat is conducted into the surrounding water layer so that the products, and the water, reach a final temperature only slightly above the initial temperature. The ∆E is the same as if the entire reaction process had occurred at the initial temperature even though the materials may have become quite hot during the course of the reaction. The heat given off by the reaction is calculated by observing the temperature rise of the water, using the condition that all heat given off by the reaction must have been absorbed by the water. Small corrections are required for the change of temperature, from the initial room temperature, of the products, and for the small amount of energy added by the hot wire or other initiation method. Some systems may also require a correction for changes of concentrations during the reaction. 7/10/07 1- 32 Problems 1. At room temperature the heat capacity (C ) of most solid elements (except the very light ones) V is about 3R. Specific heat is the heat capacity per unit mass, or the value in J/g·K. (For solids the difference between constant volume and constant pressure conditions is small and may be neglected here.) a. Find the specific heat of Pb (molar mass = 207.19). b. Find the specific heat of Cu (molar mass = 63.546). c. Find ∆E for 25 g of Cu when it is warmed from 20 C to 35 C. o o 2. Estimate the final temperature when 10 cm of iron (density 7.86 g/cm ) at 80 C is added to 30 3 3 o ml of water at 20 C. o 3. Find the work done when 3 mol of N gas at 4 atm pressure expands slowly at 25 C against a 2 o constant pressure of 1 atm. 4. Calculate the work done when 0.25 mol of SO at 27 C and 1 atm expands reversibly and 2 o isothermally to a final pressure of 0.20 atm. Find Q for the gas during this process. 5. Calculate the final temperature if an ice cube (25 g) at - 5 C is added to a cup of coffee (150 o ml, or “2/3 cup”) at 90 C, neglecting heat loss to the surroundings. o 6. The heat of vaporization of benzene, C H , is 30.72 kJ/mol and the normal boiling point is 6 6 80.1 C. o a. What is ∆E if the vapor is an ideal gas? vap b. Part of the thermal energy absorbed by a liquid as it vaporizes to a gas is required to perform work on the atmosphere as the substance expands to a vapor. What fraction of the total Q goes into work against the atmosphere when benzene (density about 0.88 g/cm ) vaporizes at 3 its normal boiling point? 7. Find ∆E at 25 C for the oxidation of S (rhombic) to give SO (gas) and for the oxidation of S o 2 to give SO (liq). The heats of reaction are - 296.81 and - 441.0 kJ/mol. Assume gases are ideal. 3 8. Calculate ∆H and ∆E for the reaction, at 25 C, o SO (g) + ½ O 6 SO (g) 2 2 3 The heat of vaporization of SO at 25 C is 43.14 kJ/mol. 3 o 9. Manganese can be prepared by a thermite process, 3 Mn O + 8 Al 6 9 Mn + 4 Al O 3 4 2 3 The standard enthalpy of formation of Mn O is - 1387.8 kJ/mol and for Al O , - 1657.7 kJ/mol. 3 4 2 3 Find the amount of thermal energy given off by the reaction as written above, starting with the reactants at room temperature and ending with the products at room temperature. 10. The interconversion of graphite and diamond does not occur at room temperature. Explain how you could determine ∆H for this transition by measurements in the laboratory. 11. Calculate the heat of vaporization of water at 50 C. o 12. Find ∆H for the hydrogenation of ethylene to produce ethane, C H + H 6 C H 2 4 2 2 6 The torr is a unit of pressure equal to 1/760 atm. It is numerically equivalent to the unit 18 “mm of Hg” but is less awkward, especially when mercury is one of several vapors present. The torr is named after Evangelista Torricelli, who invented the barometer in the 17 century. th 7/10/07 2- 33 at 150 C. The standard heats of formation, at 25 C, of ethylene and ethane are 52.4 and - 84.0 o o kJ/mol, respectively. Heat capacities (C ) of ethylene, hydrogen, and ethane within this P temperature interval are approximately 42.9, 28.8, and 52.5 J/mol·K. All three compounds are gases at room temperature and above. 13. Benzene, C H , is reduced commercially with hydrogen to give cyclohexane, C H . 6 6 6 12 C H + 3 H 6 C H 6 6 2 6 12 At room temperature (25 C) the benzene and cyclohexane are liquids and the heat of reaction is o - 205.5 kJ/mol. Benzene boils at 80.1 C with a heat of vaporization of 47.8 kJ/mol; cyclohexane o boils at 80.7 C with a heat of vaporization of 33.0 kJ/mol. The average heat capacity of liquid o benzene is 136.0 J/mol·K, of liquid cyclohexane 154.9 J/mol·K, of benzene vapor 82.4 J/mol·K, of cyclohexane vapor 150 J/mol·K, and of hydrogen 28.8 J/mol·K. Find ∆H for the reduction of benzene with hydrogen at 150 C. o [0.44 cal/g·K; 0.41; 26.7 35.8 cal/mol·K; 6.94 cal/mol·K]Check numbers 14. Air is approximately 20% O and 80% N , by volume. Find the effective molar mass of air. 2 2 Calculate the density of air at 25 C and 1 atm in kg/m , assuming it acts as an ideal gas. o 3 15. A gaseous compound, containing only sulfur and fluorine, is 62.7% (by weight) sulfur. At 27 C and 750 torr the density of the gas is 4.09 g/L. What is the molecular formula of the o 18 compound? Assume the gas is ideal. 16. Find the pressure exerted by 32 g of methane in a 2 L steel bomb at 127 C, assuming the gas o is ideal. 17. Find the pressure exerted by 32 g of methane in a 2 L steel bomb at 127 C assuming the gas o obeys van der Waals’ equation. The constants may be taken as a = 2.25 L atm/mol and b = 2 2 0.0428 L/mol. 18. Find the volume occupied by 32 g of methane at a pressure of 5 atm at 27 C if the gas is o ideal. Calculate, with this value, the n a/V correction term in the van der Waals equation and 2 2 find, from this, an approximate value for the volume occupied by the methane if it obeys van der Waals’ equation. Recalculate the correction term, with this better volume, and recalculate the volume if necessary. 19. A good vacuum obtained with a mechanical pump and a mercury diffusion pump will measure 10 torr, or better, of “noncondensable” gases (such as air), but there is 10 torr of -5 -3 mercury vapor present unless this has been trapped out. Very good vacuum systems can give 10 - torr. 9 a. How many molecules/cm are there at a pressure of 10 torr? 3 -3 b. How many molecules/cm are there at a pressure of 10 torr? 3 -9 c. What pressure would be required to achieve 1 molecule/cm ? 3 The first law prohibits certain perpetual-motion machines, called “perpetual-motion 1 machines of the first kind.” The more interesting attempts, called “perpetual-motion machines of the second kind,” do satisfy the first law. 7/10/07 2- 34 2 The Second Law of Thermodynamics During the 19 century several men, of quite different backgrounds and interests, struggled th with the basic problems of thermodynamics. Brilliant flashes of understanding were followed by years of doubting, testing, and interpreting. Among the fundamental difficulties was a confusion between two concepts. One of these was the concept of energy. The other, which is related to the general concept of equilibrium and the direction of changes with respect to equilibrium, was stated first, but the language was such that it was long misunderstood and therefore rejected. It is now accepted as the second law of thermodynamics. Very few people today who have any acquaintance with modern science would doubt the following generalizations: 1. Perpetual-motion machines don’t work. 2. Bodies in equilibrium have the same temperature. When two bodies in contact have different temperatures, energy flows, as heat (Q), from the warmer to the cooler body. 3. Bodies in equilibrium have the same pressure. When two bodies in contact, at the same temperature, have different pressures, the body at the higher pressure tends to expand and compress the body at the lower pressure. None of these statements is required by the first law, which requires energy balance in any 1 process but does not say whether a process will actually occur. For example, both exothermic and endothermic reactions are known that do proceed without external forcing. There is, despite the variety in these generalizations, a similarity of pattern and intent — saying, for example, whether a specific process will or will not occur — that suggests a common basis. The basis is found in the postulate known as the second law of thermodynamics. The Spread Function We have seen that thermodynamics largely ignores the atomic structure of matter. It is quite sufficient, for almost all purposes, to measure the macroscopic (large scale) variables such as temperature, pressure, volume, and energy. The difficulty with this approach is that it effectively hides from us one key property that may be called the “spread function,” S. Consider the “thought experiment” of adding 10 grains of yellow rice to a large bag of white rice, which will subsequently be mixed and poured and just thoroughly scrambled. Initially the yellow grains were carefully arranged, on top in one corner. Where will they be a little later after mixing? There is no energy advantage for the yellow grains to move to the bottom, or back to the top, or to any other specific location. We know that, once mixed, there is close to zero (1) T Q S rev =∆ If the ten yellow grains were regarded as indistinguishable, we should remove 10! = 3.6 x 2 10 possibilities that only interchange yellow grains among themselves. That would still leave 6 more than 10 possibilities, scarcely different qualitatively from 10 . The problem of overlapping 53 60 positions (two yellow grains being assigned to the same location) may safely be ignored at this level of probability. 7/10/07 2- 35 probability that they will spontaneously reassemble. Even if the bag of rice is only 100 grains deep, 100 grains wide, and 100 grains in the third direction, each yellow grain has 100 x 100 x 100 = 10 possible locations, which means that two grains have 10 x 10 = 10 possible locations 6 6 6 12 and the 10 grains have (10 ) = 10 possible locations. The probability they will all collect in 6 10 60 one location is not much more than 1 in 10 , which is awfully close to zero for most purposes. 60 2 Now imagine a very small number of molecules (say 10 ) to which we add 10 molecules that 6 are “excited” — perhaps they are each vibrating, or rotating, or otherwise have some extra energy that can be transferred to a neighboring molecule. After a very short time, the extra energy will be spread throughout the 10 molecules, so there is no significant chance the energy will ever 6 again (in the lifetime of the universe) be collected in the original 10 molecules. With a more typical number of molecules (such as 6 x 10 ) and many ways for each to store energy, it is not at 23 all unreasonable to say the probability of “unspreading” the energy goes quickly to zero. If we have a small steel ball (which will typically have 10 or more atoms), one way of 23 storing energy in the ball is to start it rolling. Another very large group of ways of storing energy in the ball is to let some of the atoms begin to vibrate. The mechanical energy of rolling may be converted by rolling friction to thermal energy of internal motions. There are so many possible internal motion states that we don’t expect the energy to ever spontaneously reappear as mechanical energy of rolling. Entropy Fortunately, there is an easy way to keep track of S, the spread function, without even having to count molecules. Any energy passing into an object by thermal energy transfer, Q, is “gone” in terms of having been spread around. It shows up afterward only as an increase (perhaps slight) in the temperature of the system. So the first step is to keep track of Q, and in particular, we want Q for a reversible path, or one that is as close as possible to equilibrium. Then we want to know whether this amount of energy is important to the system, or whether there is already so much energy spread around that this addition is relatively unimportant. That average energy we will find is nicely measured by the (absolute) temperature of the system. So we measure the increase in average energy as Q and divide by the temperature, T. rev Definition of Entropy Change The name was chosen by Clausius from the Greek verb entrepo (εντρεπω), meaning “to 3 turn”, implying “change”, and because it somewhat resembles the word energy, one of the important quantities involved in finding the number of states Entropy, S, is proportional to (or for suitable units, equal to) the logarithm of the relative probability, i.e., to the number of states accessible to the system. Unfortunately, it is also vaguely suggestive of enthalpy, H, which has different dimensions, units, and meaning. We also assume here that the process satisfies the first-law equation, ∆E = Q + W. 4 However, equations 1 and 5, and equation 6 (obtained below), are not subject to this restriction against other forms of work. 7/10/07 2- 36 The “official” name of the spread function, S, is entropy . By the methods of statistical 3 mechanics we can show that S is a measure of the actual number of individual states over which the energy of the system is spread. The greater the number of states (hence S), the greater the probability associated with the overall state. It is, of course, possible to transfer energy to our system without increasing the spread of energy. For example, we can transfer just enough energy, in the proper form, to make the steel ball start rolling. In a reversible process, this energy transfer is measured as work. As we saw for ideal gas expansions and compressions, the work done on the system is a minimum (and Q is a maximum) when the process is reversible. When the process is purely one of transferring energy to a mechanical mode, such as rolling or moving in a straight line, the amount of thermal energy transfer, Q = Q , is equal to zero. There is no increase in entropy. max An alternative way of approaching the second law is to recognize that ∆E is independent of path, so we can write, for any process at constant temperature, between fixed end points 4 (T) ∆E = Q + W = Q + W = Q + W (2) rev rev max min There can be only one value of Q and one value of W , so we can represent these, also, as max min values “independent of path”, which for the moment we will label as ∆A (= W ) and as ∆B (= rev Q ) and regard A and B as state functions. rev (T) ∆E = ∆A + ∆B or E = A + B (3) In any isothermal process in which the energy of the system changes by ∆E, some part of this energy change, equal to ∆A, must appear as work done on the system. The second energy term, called ∆B for the moment, cannot be lost, because of the first law (conservation of energy); however, it may be regarded as “spilled energy”, which is lost for useful purposes during the transfer operation because it has become spread through the system. From equation 3, two important functions are obtained. One, temporarily given the symbol A, is called the Helmholtz free energy, or sometimes the work function, and is characterized, as above, by the constant temperature equation [...]... metal, well insulated from its surroundings and large enough so that the reasonable amount of thermal energy added or withdrawn will not change its temperature For the ice, ∆S = Q rev 33 3 6 J/g = = 1 22 J/g ⋅ K 2 73 15 K T If the thermostat is at exactly the same temperature, ∆Sthermostat = − 33 3.6 J/g (ice) = −1.22 J/g ⋅ K 2 73. 15 K Taking the ice as the system and the thermostat as the surroundings, (∆S)... and the ice, the ice will not melt We cannot melt an ice cube by bringing it into contact with a large block of ice; it must be touched to something warmer If the temperature of the thermostat is higher than 2 73. 15 K, ∆Sthermostat will be smaller in magnitude than ∆Sice , and ∆S for the ice plus the thermostat, or system plus surroundings, will be positive This is the essence of the second law of thermodynamics, ... q rev T (5) Statistical mechanics tells us something about the value of S, representing the number of molecular states, but thermodynamics (which never quite admits the importance of molecules) tells us only how to find changes in entropy, S ENTROPY, EQUILIBRIUM, AND SPONTANEOUS CHANGE Consider the process of melting 1 g of ice, at 0oC and 1 atm pressure, by bringing it into contact with a thermostat,... there is a change of chemical composition, as would be required if electrical work were produced, or a change of concentrations, additional terms should be added to the first-law equation and to equation 7.) At constant volume the second term of equation 7 becomes zero and we obtain or  ∂E    =T  ∂S V 1  ∂S    =  ∂E V T 7/10/07 2- 38 (8) (8a) Take an arbitrary system, at constant volume,... its surroundings (so that Q = 0 and ∆E = 0) The second law then requires that dS $ 0 for any changes within this system Divide the system into two parts in any fashion (Figure 1) For example, the total system might be a cup of water, which is considered to be divided into two equal parts, or it might be a bar of lead, with one cm3 in the center considered as one part and the remainder as the second... 7/10/07 2- 37 (∆S)system + surroundings > 0 (6) for any process whatever that actually occurs The similarity, and contrast, to the statement of the first law (equation 2, Chapter 1) should be noted In the limit of a process that is truly reversible (for example, thermostat and ice at the same temperature, so that the ice can melt or refreeze by exchange of thermal energy with the thermostat), the total... are, in general, many different paths that a system may follow between any two specified states, of which at least one (actually, an infinite number) will be thermodynamically reversible Let q i and w i be the amount of thermal energy transferred and the amount of work done along the i th path, or more generally, q 1, q 2, ··· q rev are the various amounts of thermal energy transferred and w 1, w 2, ···... more current notation, F), and an even more useful quantity derived from it, will be investigated later The spread function, S, is obtained from the second function, B, when we divide by the temperature, T That is, S = B/T or B = TS, where S is the spread function, entropy, previously introduced Then a small change in B breaks into two parts d(TS) = T dS + S dT and at constant temperature (T) d(TS)... …qrev and w1 …w2 … ··· …wrev But dE is the same for each of these paths That is, regardless of which path the system actually follows, dE = q1 + w1 = q2 + w2 = ··· = qrev + wrev and, because qrev = T dS and wrev = - P dV (if the only work is work of expansion or compression), it follows that, for any path, dE = T dS - P dV (7) This equation is valid for any process occurring in a system of constant composition... one part and the remainder as the second part Let E1 be the energy of the first part and E2 the energy of the second part, and consider a flow of energy (as thermal energy transfer, Q) from one part to the other, holding the two sub-system volumes constant Then dE = dE1 + dE2 = 0 dE1 = - dE2 (9) The total entropy change, dS, arising from a flow of energy between the two parts, is dS = (E, V) = dS dS . ½ O 6 SO (g) 2 2 3 The heat of vaporization of SO at 25 C is 43. 14 kJ/mol. 3 o 9. Manganese can be prepared by a thermite process, 3 Mn O + 8 Al 6 9 Mn + 4 Al O 3 4 2 3 The standard enthalpy of. atoms), one way of 23 storing energy in the ball is to start it rolling. Another very large group of ways of storing energy in the ball is to let some of the atoms begin to vibrate. The mechanical. as above, by the constant temperature equation (5) T q dS rev = KJ/g 22.1 K 15.2 73 J/g 6 .33 3 ⋅===∆ T Q S rev KJ/g 22.1 K 15.2 73 (ice) J/g 6 .33 3 ⋅−= − =∆ thermostat S 7/10/07 2- 37 (T) ∆A = W (4) rev This